Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 3 (Q.No. 45)
45.
A mild steel bar is in three parts, each 20 cm long. The diameters of parts AB, BC and CD are 2 cm, 1 cm and 3 cm respectively.
The bar is subjected to an axial pull of 4t as shown in the given figure. If E = 2 x 106 kg/cm2 and the elongations in the three parts of the bar are Δ1, Δ2 and Δ3 respectively, then the ratio of the greatest to the least of these elongations will be
Discussion:
6 comments Page 1 of 1.
Chhaya said:
8 years ago
Answer should be A.
Because elongation is inversely proportional to Area here is given diameter.
Because elongation is inversely proportional to Area here is given diameter.
Sujatha said:
5 years ago
Please explain this.
Rocky bhai said:
4 years ago
Yes, A is the right answer.
ABHI said:
4 years ago
Thanks for explaining.
(1)
Wakeel said:
2 years ago
Elongation is inversely proportional to area.
@All
Here, Most of us will solve this question considering the bars as rectangular and will come out with an answer = 3, which is wrong.
You have to consider these bars as circular and you will get an answer =9, which is correct.
@All
Here, Most of us will solve this question considering the bars as rectangular and will come out with an answer = 3, which is wrong.
You have to consider these bars as circular and you will get an answer =9, which is correct.
(2)
Abhishek said:
2 years ago
∆ = PL/AE.
Three of these are constants P,L,E, except area which is πd²/4, from what we conclude that ∆ = 1/d²,
So, compare the greatest and least elongation,
(1/1² ÷ 1/3²) = 9.
Three of these are constants P,L,E, except area which is πd²/4, from what we conclude that ∆ = 1/d²,
So, compare the greatest and least elongation,
(1/1² ÷ 1/3²) = 9.
(5)
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